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Neporo4naja [7]
3 years ago
14

A train has an initial velocity of 44m/s and an accelaration of _4m/s calculate its velocity​

Physics
1 answer:
Kobotan [32]3 years ago
7 0

Complete question:

A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity​ after 10s ?

Answer:

the final velocity of the train is 4 m/s.

Explanation:

Given;

initial velocity of the train, u = 44 m/s

acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)

time of motion, t = 10 s

let the final velocity of the train = v

The final velocity of the train is calculated using the following kinematic equation;

v = u + at

v = 44 + (-4 x 10)

v = 44 - 40

v = 4 m/s

Therefore, the final velocity of the train is 4 m/s.

You might be interested in
-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a
bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1

The velocity of the carts after the event is 1 m/s

3 0
2 years ago
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F
LiRa [457]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

8 0
3 years ago
When will you say a body is in a) uniform acceleration (b) non uniform acceleration
s2008m [1.1K]

Answer:

See below ~

Explanation:

Part (a) :

We can say a body is in uniform acceleration if the acceleration of the object remains constant with respect to time throughout its motion.

Part (b) :

We can say a body is non-uniform acceleration if the acceleration of the body varies with respect to time throughout its motion.

7 0
2 years ago
A housekeeper is cleaning the kitchen and decides to to use a ladder to reach a top shelf. As he/she is cleaning he/she is start
pshichka [43]

The amount of gravitational force between both objects will be the same.

The magnitude of the Earth's gravitational force exerted on the housekeeper is calculated by applying Newton's second law of motion;

F = mg

where;

<em>m </em><em>is the mass of the housekeeper</em>

<em>g </em><em>is acceleration due to gravity</em>

According to Newton's third law of motion, action and reaction are equal and opposite.

The force exerted on the housekeeper by the Earth is equal in magnitude to the force exerted on the Earth by the housekeeper.

F_{H} = - F_{E}

The two forces are equal in magnitude but opposite in direction.

Thus, the correct option is " the amount of gravitational force between both objects will be the same"

<em>The</em><em> missing part</em><em> of the </em><em>question </em><em>is below:</em>

a. the Earth exerts the largest amount of gravitational force

b. the housekeeper exerts the largest amount of gravitational force

c. the amount of gravitational force between both objects will be the same

Learn more about Newton's third law of motion here: brainly.com/question/15507

3 0
2 years ago
Would earth orbiting the sun be known as
Nitella [24]
C. Newton’s Third Law of Motion.

Because...
Newtons third law implies conversation of momentum it can also be seen as following from the second law: when one object pushes a second object at some point of contact using an applied force, there must be an equation of opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.
5 0
3 years ago
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